Derivation of the third Kinematic Equation

13

To start off, we have the two previous equations to work with:

$$ \Delta x = v_0 t + \frac{1}{2} a t^2 $$
$$ v = v_0 + a t $$

Our goal is to derive the third equation:

$$ v^2 = v_0^2 + 2 a \Delta x $$

Note that this new equation has no $%t$% component, which is what makes it useful. It also signifies that we'll have to remove the $%t$% somehow. To do that, we can use substitution. First, let's solve for $%t$%:

$$ t = \frac{v - v_0}{a} $$

We also need to solve for $%t^2$%, because of the occurrence of it in the first equation:

$$ t^2 = (\frac{v - v_0}{a})^2 $$
$$ t^2 = \frac{v^2 - 2 v_0v + v_0^2}{a^2} $$

With this, we can do a simple substitution, and decompose some of the resultant polynomials so that we can cancel some of the terms. Namely, we'll split the $%v - v_0$% terms into two separate fractions, and we'll also cancel one of the $%a$%'s under the fraction:

$$ \Delta x = v_0 (\frac{v - v_0}{a}) + \frac{1}{2} a (\frac{v^2 - 2 v_0v + v_0^2}{a^2}) $$
$$ \Delta x = \frac{v_0 v}{a} - \frac{v_0^2}{a} + \frac{v^2}{2 a} - \frac{v_0v}{a} + \frac{v_0^2}{2a} $$

After canceling the $%v_0 v$%'s and grouping the like terms, we get:

$$ \Delta x = \frac{v^2}{2 a} - \frac{v_0^2}{2 a} $$

Multiply both sides by $%2a$% and solve for $%v^2$%, and you get the equation that we were looking for:

$$ 2 a \Delta x = v^2 - v_0^2 $$
$$ v^2 = v_0^2 + 2 a \Delta x $$

asked 02 Jul '12, 22:47

Zaven%20Muradyan-1's gravatar image

Zaven Murady...
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accept rate: 79%


11 Answers:

12next »

I think there is an easier derivation. We start with $% v = v_0 + a t $% and squaring both sides, we get:

$$ v^2 = v_0^2 + a^2 t^2 + 2v_0at $$
$$ \Rightarrow v^2 = v_0^2 + 2a (v_0t + \frac{1}{2}at^2) $$
$$ \Rightarrow v^2 = v_0^2 + 2a\Delta x.$$

The last line comes by using $% \Delta x = v_0t + \frac{1}{2}at^2. $%

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answered 02 Jul '12, 23:23

ked4r-4's gravatar image

ked4r-4
8.9k737

1

Aha! That's much nicer. I knew there must have been an easier way! Thanks!

(02 Jul '12, 23:30)

Zaven Murady...

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@ked4r: Oh god! I'm lost. are you using integral or derivation? It looks like an integral.

(12 Jul '12, 13:15)

Yulianto

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@Yulianto: no integral or derivation involved here. Take the first line, v = v0 + a*t, and just take the square of it.

(12 Jul '12, 13:19)

Radoslaw Jurga

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@radoslaw-jurga: Hmm.. I know this math. is it completing square?

(12 Jul '12, 14:14)

Yulianto

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@Yuilanto I am just squaring both sides of v = v0 + at. Squaring the left side give v^2. Squaring the right side gives (v0 + at) * (v0 + at) = v0 * (v0 + at) + at * (v0 + at) = v0^2 + a^2t^2 + 2v0 at.

(12 Jul '12, 20:55)

ked4r-4

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Once I realized about eliminating the t I was able to do this and am really shocked and proud of myself. I was totally terrified of the challange and had seriously considered skipping it.

Perhaps it might be nice to have an instructor comment giving that one very basic hint.

(16 Aug '12, 06:49)

Jessica Hyde

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It took me a little while to derive the third equation, but I finally was able to! Looks like I had the same idea as Zaven, which still feels more intuitive to me than the method described by ked4r-4. I guess only because I wouldn't casually square an equation just so I could fiddle with it some more after. ; )
But I am impressed and I definitely enjoy seeing the different ways in which people solve these problems!

(08 Nov '13, 13:05)

Tanya-42

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Ok, bear with me a little. How do you get from the first to the second equation (after squaring both sides)? Are you factorizing $% a^2t^2+2v_0at $% ?

(03 Dec '13, 13:50)

Ashilur

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Using this equation, show that if h is the maximum height that an object rises when given a vertical velocity that Vo=sqrt2gh

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answered 26 Oct '13, 00:33

Maria-190's gravatar image

Maria-190
11

I'm still confused why the average velocity is used when deriving the equation for the displacement. Supposing that we have initial velocity = 0 and uniform acceleration = a, then we get the following table:

           S|   V|T
           0|  0a|0
          1a|  1a|1              
     1a + 2a|  2a|2
1a + 2a + 3a|  3a|3

...

 1a +...+ ta|  ta|t

where T = current time,V = velocity at time T, S = displacement at time T

This gives $% S = \sum_{i=1}^{t}ia = \frac{at}{2}(t + 1) $% for the displacement.
What am I getting wrong here?

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answered 12 Jul '12, 20:48

Mila%20Grigorova's gravatar image

Mila Grigorova
21

edited 12 Jul '12, 20:57

You can try out the same exercise using time increments of 0.5 instead of 1. In that case you will find the expression to be S = 0.5 a sum{i=1 to 2T} i = aT(T+0.5)/2. In general as you keep decreasing the time increments to 0.25, 0.125, etc. you will find the general expression aT(T+dt)/2. As dt goes to zero (in the limit) the displacement will be given by aT^2/2. The key is that velocity is continuously increasing rather than in a step-wise fashion as obtained from your table.

(13 Jul '12, 01:35)

ked4r-4

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Hello Mila, your table is not correct for the displacement. Notice that, for example, in the time interval from t0=0 to t1=1 s, velocity changed from vo=0 to v1=0+1.a=1.a! So the displacement in this interval is the average velocity in this interval: v_m(0;1)=(v(0)+v(1))/2 times the time interval 1-0=1, THAT IS: S(0;1)=v_m(0;1). (1-0)=(0+1.a)/2 . 1=1/2 . a . 1^2.

Notice that if you use only the final velocity, in your calculation of the displacement, you will obtain a greater displacement that it was in reality since the body only has that velocity at the end of the interval.

On the other hand, if you only had used the initial velocity, in your calculation of the displacement, you would had obtained a lesser displacement that it was in reality since the body starts with zero velocity but it increases continually IN the time interval.

In conclusion: you can not use only one velocity, may that be the initial or the final, in your calculation of the displacement in a given time interval. You have to use the average velocity!

Hope its useful.

(13 Jul '12, 07:15)

Rosa Brigida... ♦

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In the above table you have drawn, the column for S does not contain correct values. E.g. If the velocity at T=2 is 2a, it does not mean that S would be 1a+2a because:

the acceleration is not increasing step-wise, it is a continuous function. And also displacement = velocity*time, and in the table you have drawn there is now time factor in the values in column S. Displacement will be a function of the square of time because we get acceleration when we differentiate a function of displacement 'twice'.

Hence there lies your mistake.

Hope this explains!

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answered 13 Jul '12, 01:18

Pranav%20Kumar's gravatar image

Pranav Kumar
4156

to derive all equations consider this: initial velocity=u and final velocity is v.

s=1/2((v+u)/t) --- equation 1 (as displacement is average velocity * time (here we are considering displacement in an interval))

put v=u+at in equation 1 to get,

s=0.5(u+u+at)t which is same as s=ut+0.5at^2----equation 2

to derive the next equation, observe that we need to eliminate time factor t.

so in equation 2 put t as (v-u)/t. then we will get:

x=u(v-u)/a + 0.5a*((v-u)/a)^2) expand the square term to solve and get the 3rd equation of motion.

hope this explains!

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answered 12 Jul '12, 13:45

Pranav%20Kumar's gravatar image

Pranav Kumar
4156

Question: How do you guys use the proper math symbols on your computer?

Anyway, here is my derivation:

delta x = dx
v zero = v0

Starting equations:

1) dx = v0t + 0.5at^2
2) v = v0 + at
3) a = (v - v0)/t

Rearranging equation 3 to solve for t:

4) t = (v - v0)/a

Substituting for equation 3 for 'a' in equation 1:

dx = v0t + 0.5(v - v0)t

dx = 0.5v0t + 0.5vt

2dx = v0t + vt = (v0 + v)t

Substituting equation 4 for t:

2dx = (v + v0)(v - v0)/a

2adx = (v + v0)(v - v0)

2adx = v^2 - v0^2

v^2 = v0^2 + 2adx

Therefore, the solution is:

v^2 = v0^2 + 2adx

Question: How do you guys use the proper math symbols on your computer?

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answered 10 Jul '12, 07:00

Roman%20Lajciak's gravatar image

Roman Lajciak
1.8k101937

(10 Jul '12, 07:23)

Radoslaw Jurga

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Thanks a lot!

(10 Jul '12, 18:13)

Roman Lajciak

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Here is what Peter Norvig would call a back of an envelope derivation... Haha:

Third Kinematic Equation Derivation

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answered 06 Jul '12, 03:44

Jardi%20Martinez's gravatar image

Jardi Martinez
2.1k21037

edited 06 Jul '12, 03:47

Yet another way to get there for people a little bit ahead and that know more maths:

$$ \frac{dv}{dt} = a $$

$$ v \frac{dv}{dt} = av $$

$$ \frac{d}{dt} \left( \frac{v^2}{2} \right) = av = a \frac{dx}{dt} $$

$$ d \left( \frac{v^2}{2} \right) = a dx $$

$$ \int d \left( \frac{v^2}{2} \right) = a \int dx $$

$$ \frac{v^2}{2} - \frac{v^2_0}{2}= a (x - x_0) $$

$$ v^2 = v^2_0 + 2a(x - x_0) $$

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answered 03 Jul '12, 03:53

Radoslaw%20Jurga's gravatar image

Radoslaw Jurga
6.9k41237

How did v(dv/dt) = av become (d/dt)(v^2/2) = av? Why do we divide v^2 by 2?

(05 Jul '12, 12:54)

Douglas Browne

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The derivative of v^2 is 2 * v * dv/dt. (We use the rule for powers and the chain rule) You see that there is a factor 2, so v * dv/dt is half of the derivative of v^2.

(05 Jul '12, 13:58)

Radoslaw Jurga

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Ah didn't see that, thanks.

(05 Jul '12, 18:48)

Douglas Browne

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There's another way to deduce this third equation, using the kinetic theorem (total variation of energy is equal to total energy transfer), the second law of Newton (resultant force is directly proportional to the mass and to the acceleration acquired by the particle) and the definition of work of a force along a displacement (infinite sum of the infinitesimal works along infinitesimal displacements). For a particular case of a constant resultant force in the x direction: $$ \Delta E_c (t_o;t)=\int_{t_o}^t (\vec{F}_R \:\dot \: d \vec{r})$$ $$\frac{1}{2}m (v^2-v_o^2)=m \: \vec{a} \: \dot \: \Delta \vec{r}$$ $$ v^2=v_o^2 +2 \: \vec{a} \: \dot \: \Delta \vec{r}$$ $$ v^2=v_o^2 +2 \: a_x \: \Delta x$$.

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answered 04 Jul '12, 10:57

Rosa%20Brigida%20Fernandes's gravatar image

Rosa Brigida... ♦
12.2k213

edited 04 Jul '12, 11:02

What is the third kinematic equation calculating? What would it be useful for?

Couldn't you use either of the first two equations to solve for anything that the third equation can?

-Thanks

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answered 03 Jul '12, 18:01

Josh%20Davoll's gravatar image

Josh Davoll
3211113

2

Yes you could, you could solve for the time in one equation, and plug the time in the other equation. But it's slower and more tedious than just using this one.

(03 Jul '12, 18:04)

Radoslaw Jurga

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5

It's useful for problems where you don't need to solve for time, but want to relate velocity, acceleration, and displacement. Technically, though, the equation gives you no additional power beyond the first two.

(03 Jul '12, 18:05)

Jonathan Burket ♦♦

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Thanks very much :)

(03 Jul '12, 18:49)

Josh Davoll

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Asked: 02 Jul '12, 22:47

Seen: 4,571 times

Last updated: 03 Dec '13, 13:52